Every Linear Order Isomorphic to Its Cube Is Isomorphic to Its Square
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EVERY LINEAR ORDER ISOMORPHIC TO ITS CUBE IS ISOMORPHIC TO ITS SQUARE GARRETT ERVIN Abstract. In 1958, Sierpi´nskiasked whether there exists a linear order X that is isomorphic to its lexicographically ordered cube but is not isomorphic to its square. The main result of this paper is that the answer is negative. More generally, if X is isomorphic to any one of its finite powers Xn, n > 1, it is isomorphic to all of them. 1. Introduction Suppose that (C; ×) is a class of structures equipped with a product. In many cases it is possible to find an infinite structure X 2 C that is isomorphic to its own square. If X2 =∼ X, then by multiplying again we have X3 =∼ X. Determining whether the converse holds for a given class C, that is, whether X3 =∼ X implies X2 =∼ X for all X 2 C, is called the cube problem for C. If the cube problem for C has a positive answer, then C is said to have the cube property. The cube problem is related to three other basic problems concerning the mul- tiplication of structures in a given class (C; ×). 1. Does A × Y =∼ X and B × X =∼ Y imply X =∼ Y for all A; B; X; Y 2 C? Equivalently, does A × B × X =∼ X imply B × X =∼ X for all A; B; X 2 C? 2. Does X2 =∼ Y 2 imply X =∼ Y for all X; Y 2 C? 3. Does A × Y =∼ X and A × X =∼ Y imply X =∼ Y for all A; X; Y 2 C? Equivalently, does A2 × X =∼ X imply A × X =∼ X for all A; X 2 C? The first question is called the Schroeder-Bernstein problem for C, and if it has a positive answer, then C is said to have the Schroeder-Bernstein property. The second question is called the unique square root problem for C, and if its answer is positive, then C has the unique square root property. Taken together, the first two questions are sometimes called the Kaplansky test problems, after Irving Kaplansky who posed them in [7] as a heuristic test for whether a given class of abelian groups (under the direct product) has a satisfactory structure theory (\I believe their defeat is convincing evidence that no reasonable invariants exist"). Tarski [14] had posed them previously for the class of Boolean algebras. All three questions are listed in Hanf's seminal paper [4] on products of Boolean algebras. We will refer to Question 3 as the weak Schroeder-Bernstein problem for C, and the corresponding property as the weak Schroeder-Bernstein property. A negative solution to Question 3 obviously gives a negative solution to Question 1. If the product for C is commutative, it gives a negative solution to Question 2 as well. If the cube problem for C has a negative solution, that is, if there is an X 2 C that is isomorphic to its cube but not to its square, then all three questions have a negative solution, even without assuming commutativity of the product. In practice, it is often by constructing such an X that these three problems are solved. 1 2 GARRETT ERVIN If the class C does not contain any infinite structure isomorphic to its cube, then the cube property holds trivially. When the cube property does not hold trivially, it usually fails. The first result in this direction is due to Hanf, who constructed in [4] a Boolean algebra that is isomorphic to its cube but not its square. Tarski [15] and J´onsson[6] immediately adapted Hanf's result to show the failure of the cube property for the class of semigroups, the class of groups, the class of rings, and various other classes of algebraic structures. Hanf's example, and consequently many of those produced by Tarski and J´onsson,is of size continuum, and for some time it was open whether there were countable examples witnessing the failure of the cube property for these various classes. In 1965, Corner showed in [1] that indeed there exists a countable (torsion-free, abelian) group G isomorphic to G3 but not G2. Later, Jones [5] showed that it is even possible construct a finitely generated (necessarily non-abelian) group isomorphic to its cube but not its square. In 1979, Ketonen [8] solved the so-called Tarski cube problem by producing a countable Boolean algebra isomorphic to its cube but not its square. Throughout the 1970s and 1980s, Trnkov´asolved the cube problem negatively for many different classes of topological spaces and relational structures, including the class of graphs under several different notions of graph product [19]. Her topo- logical results are summarized in [18]. Answering a question of Trnkov´a,Orsatti and Rodino showed in [11] that there is even a connected topological space home- omorphic to its cube but not its square. Koubek, Neˇsetˇril,and R¨odl[9] showed that the cube property fails for the class of partial orders, as well as for other classes of relational structures. More recently, Eklof and Shelah [2] constructed an @1-separable group isomorphic to its cube but not its square, and Gowers [3] constructed a Banach space linearly homeomorphic to its cube but not its square. On the other hand, there are rare instances when the cube property holds non- trivially. It holds for the class of sets under the cartesian product, since any set in bijective correspondence with its cube is either infinite, empty, or a singleton, and hence in bijection with its square. This is immediate if one assumes the axiom of choice, but it can be proved without the axiom of choice using the Schroeder- Bernstein theorem. Similarly easily, the cube property holds for the class of vector spaces (over a fixed field) under the direct product. Less trivially, the cube prop- erty holds for the class of countably complete Boolean algebras. This follows from the Schroeder-Bernstein theorem for such algebras. Trnkov´a[16] showed that the cube property also holds for the class of countable metric spaces (where isomor- phism means homeomorphism), as well as for closed subspaces of Cantor space [17]. Koubek, Neˇsetˇril,and R¨odlshowed in [9] that the cube property holds for the class of equivalence relations. It is worth noting that for all of these classes, it is actually possible to establish the stronger Schroeder-Bernstein property. In his 1958 book Cardinal and Ordinal Numbers [13], Sierpi´nskiposed the cube problem (although he does not use the term) for the class (LO; ×lex) of linear orders under the lexicographical product. On page 232, he writes, \We do not know so far ::: any type α such that α = α3 6= α2." Here, \type" means linear order type, and the ordering on the cartesian powers α2 and α3 is the lexicographical ordering1. Although the cube problem has been solved 1Sierpi´nskiactually ordered these powers anti-lexicographically, though in this paper we will use the lexicographical ordering. This does not change the problem. THE CUBE PROPERTY FOR THE CLASS OF LINEAR ORDERS 3 for many other classes of structures, Sierpi´nski'squestion has remained open. One major difference in this version of the cube problem is that, unlike the products for the other classes so far discussed, the lexicographical product of linear orders is not commutative. Though he does not make a conjecture in his book, his language suggests that Sierpi´nskiexpected that such an α exists, that is, that the cube property fails for (LO; ×lex). He was already aware of examples of linear orders witnessing the failure of the unique square root property and (the right-sided and left-sided versions of) the Schroeder-Bernstein property. The main result of this paper is that in fact the cube property holds for (LO; ×lex). This appears as Theorem 4.14 below. Main Theorem. If X is a linear order and X3 =∼ X, then X2 =∼ X. More generally, if Xn =∼ X for some n > 1, then X2 =∼ X. Thus the cube property holds for the class of linear orders despite the fact that the Schroeder-Bernstein property and unique square root property fail. We will show in Section 6 that even the weak Schroeder-Bernstein property fails for (LO; ×lex). In this sense, the cube property is closer to failing for (LO; ×lex) than it is for the other classes for which it is known to hold. In the remainder of this paper we prove the main theorem as well as some related results. In Section 2, we define the necessary notation and terminology, and give some examples of countable linear orders isomorphic to their own squares. We also give the easy proof that the cube property holds for linear orders with both a top and bottom point. In Section 3, we prove a representation theorem for linear orders X that are invariant under left multiplication by a fixed order A, that is, that satisfy the isomorphism A×X =∼ X. It turns out that such orders can essentially be represented as unions of tail-equivalence classes in the lexicographically ordered tree A!. This theorem can be generalized to characterize the isomorphism An × X =∼ X for any n ≥ 1, and in particular A2 × X =∼ X. It is then possible to write down a sufficient condition, namely the existence of a parity-reversing automorphism of A!, for the implication A2 × X =∼ X =) A × X =∼ X to hold for every X. The combinatorial heart of the proof is contained in Section 4, where parity- reversing automorphisms of A! are constructed for various orders A, including all countable orders.